91Ó°ÊÓ

The goal is to use the following formulas to compute $\mathrm{SST},\mathrm{SSTR}$, and $\mathrm{SSE}$:

$SST=\underset{i=1}{\overset{n}{∑}}{\left(X_i-\overline{X}\right)}^2$denotes the entire sum of squares.

Where,

$\stackrel{¯}{X}=\frac{{\displaystyle \underset{i=1}{\overset{n}{∑}}}{X}_i}{n}$

Calculate the average:

$$\begin{gathered} \stackrel{¯}{X}=\frac{{\displaystyle \underset{i=1}{\overset{n}{∑}}}{X}_i}{n} \\ =\frac{(4+8+9+...+2+9)}{15} \\ =\frac{75}{15} \\ =5 \end{gathered}$$

Determine the SST,

$$\begin{gathered} SST=\underset{i=1}{\overset{n}{∑}}{(X_i-\overline{X})}^2 \\ ={(4-5)}^2+{(8-5)}^2+{(9-5)}^2+...+{(2-5)}^2+{(9-5)}^2 \\ =112 \end{gathered}$$

Thus,

$SST=112$

b.

Determine the treatment mean square $\left(\mathrm{SSTR}\right)$:

$SSTR=\underset{i=1}{\overset{n}{∑}}{n}_i{({\overline{X}}_i-\overline{X})}^2$

$=n_1{\left({\stackrel{¯}{X}}_1-\stackrel{¯}{X}\right)}^2+n_2{\left({\stackrel{¯}{X}}_2-\stackrel{¯}{X}\right)}^2+n_3{\left({\stackrel{¯}{X}}_3-\stackrel{¯}{X}\right)}^2+n_4{\left({\stackrel{¯}{X}}_4-\stackrel{¯}{X}\right)}^2+n_5\left({\stackrel{¯}{X}}_5-\stackrel{¯}{X}\right)$

localid="1653227695948" $$\begin{gathered} =3{(3-5)}^2+3{(6-5)}^2+3{(8-5)}^2+3{(2-5)}^2+3{(6-5)}^2=3[4+1+9+9+1] \\ =3×24 \\ =72 \\ SSTR=72 \end{gathered}$$

$\mathrm{SSE}$Error Mean Square $=\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2$

The following is the generic formula for variance $S_i^2$

$s_i^2=\frac{1}{n_i-1}\underset{i=1}{\overset{n}{∑}}{\left(X_i-{\stackrel{¯}{X}}_i\right)}^2$

The following is the MINITAB output for variances:

Therefore,

$$\begin{gathered} SSE==\left(n_1-1\right)s_1^2+\left(n_2-1\right)s_2^2+\left(n_3-1\right)s_3^2+\left(n_4-1\right)s_4^2+(n_5-1)s_5^2 \\ =(3-1)1+(3-1)3+(3-1)3+(3-1)4+(3-1)9 \\ =2×20 \\ =40 \end{gathered}$$

Hence

$SSE=40$

c.

There is no significant difference in the sample means, according to ${\mathrm{H}}_0$.

There is a significant discrepancy in the sample means ${\mathrm{H}}_1$.

The following is the technique for performing One-Way Analysis of Variance:

1) Open the $\mathrm{MINITAB}$sheet and import the data.

2) Select $\mathrm{Start}→\mathrm{ANOVA}→\mathrm{One}-\mathrm{way}$(un-stacked),

3) In the$\mathrm{Response}\mathrm{text}\mathrm{box}$, type $\mathrm{RESPONSE}$.

4) Enter $95$as the confidence level.

5) Finally, press the $\mathrm{OK}$button.

The following is the output:

d.

$F=4.50$is the determined value. $p=0.024$is the value. As a result, $\mathrm{p}$has a value less than$alpha=0.05$. Therefore, the Null hypothesis must be rejected. As a result, draw the conclusion that the sample means differ significantly.